Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7
and target 7
,
A solution set is:
[7]
[2, 2, 3]
class Solution {
public:
vector<vector<
int> > combinationSum(vector<
int> &candidates,
int target)
{
//
sort candidates
int size=candidates.size();
for(
int i=
0;i<size;i++)
for(
int j=i+
1;j<size;j++)
if(candidates[i]>candidates[j])
{
int tmp=candidates[i];
candidates[i]=candidates[j];
candidates[j]=tmp;
}
//
dfs
vector<vector<
int>> result;
if(size==
0)
return result;
int v[target/candidates[
0]];
int cdep=
0;
int vdep=
0;
dfs(result,candidates,size,v,cdep,vdep,target);
return result;
}
void dfs(vector<vector<
int>>& result,vector<
int> &candidates,
int& csize,
int* v,
int& cdep,
int& vdep,
int& target)
{
if(target==
0)
{
vector<
int> vals;
for(
int i=
0;i<vdep;i++)
vals.push_back(v[i]);
result.push_back(vals);
return;
}
if(cdep==csize || target<candidates[cdep])
return;
int dup=target/candidates[cdep];
for(
int i=
0;i<=dup;i++)
{
//
add to v
for(
int j=
0;j<i;j++) v[vdep+j]=candidates[cdep];
//
search next candidate
target=target-i*candidates[cdep];
cdep=cdep+
1;
vdep=vdep+i;
dfs(result,candidates,csize,v,cdep,vdep,target);
vdep=vdep-i;
cdep=cdep-
1;
target=target+i*candidates[cdep];
}
}
};